Kingpin and Bogie Load Split on a Semitrailer

There is no standard for this and none is needed. Every other empirical relation in this shard carries a fitted coefficient, a calibration and a caveat. This one is a simply supported beam: the trailer rests on the kingpin at the front and its bogie at the back, the sum of moments about either support is zero, and the reactions follow. It is first-year statics, it is exact, and it is the most trustworthy page here.

Take moments about the centre of the trailer axle group and the kingpin reaction falls straight out: $R_{kp} L = W(L-a)$, so $R_{kp} = W(L-a)/L$. The axle group takes the remainder, $W a / L$. The lever is the whole story — the further back the centre of gravity sits, the more the trailer's own wheels carry and the less goes to the tractor.

Why this is called a two-stage problem. The kingpin reaction does not land on the drive axles. It arrives at the fifth wheel, which sits somewhere between the tractor's steer axle and its drive tandem, and so the same lever argument runs a second time on the tractor to split $R_{kp}$ between steer and drives. Move the fifth wheel forward and more of the trailer's weight lands on the steer axle; slide it back and the drives take it. Two moment balances, done in order, and the complete five-axle weight distribution follows from four dimensions and two weights. Every tractor has a sliding fifth wheel for exactly this reason, and it is the coarse adjustment where the trailer's slider is the fine one.

This is the arithmetic behind every trailer bogie slid at a weigh station. Sliding the trailer tandem changes $L$. Forward — smaller $L$, shorter lever — puts more weight on the trailer axles and less on the kingpin, which takes weight off the drives. Backward does the reverse. A driver who comes off the scale with heavy drives and light trailer axles slides the tandem back; heavy trailer axles and light drives, slide it forward. The calculation takes a minute and is far better done in the yard than in the queue.

Two things the equation cannot see, and both bite. The slider does not move continuously — it locks into pin holes at a fixed pitch, typically six inches — so the answer has to be rounded to a hole and rechecked. And more importantly, moving the bogie changes the axle SPACING as well as the load, which moves the bridge-formula limit for every group that includes those axles. Fixing a drive-axle overload by sliding the tandem forward shortens the steer-to-rear-axle span and can put the vehicle over the bridge formula on a group that was legal a moment ago. The two checks have to be done together. Several jurisdictions also cap how far back the trailer tandem may legally be set regardless of weight, because a long rear overhang makes the trailer off-track and sweep wide through turns.

One case the arrangement above deliberately excludes: a centre of gravity behind the trailer axle group. That puts the kingpin into tension — the fifth wheel is being lifted rather than loaded — and the tractor's steering goes light. It is a real and dangerous loading condition, it happens when a trailer is loaded from the back forward and stopped early, and it is outside what this form of the equation describes. The fix is not arithmetic; it is to move the load forward.

The last direction is the most useful of the four. The centre of gravity of a loaded trailer is the one quantity nobody can measure directly — and two scale readings determine it exactly. Weigh the axle groups, put the numbers into the rearrangement for $a$, and you know where the load actually sits. Compare that against where the loading plan said it would sit, and the difference is your answer to why the drives are heavy.